System identification based on Step and Impulse response considering first and second order transfer function models

These notes discuss system identification based on the step and impulse response. In the following we consider linear, time-invariant systems of first and second order, as they provide reasonable approximation for the description of the dominant behavior of most linear time-invariant systems. I. First order system A first order system is described in frequency domain by the transfer function k G (s) = t s +1 where the parameters k and t are the system gain ( k ) and the time constant of the system ( t ). The step response of this system, considering initial conditions equal with 0, is obtained as follows: a. the Laplace transform of the system input is 1 U (s) = s b. using the relation Y ( s ) = H ( s )U ( s ) one obtains the Laplace transform of the output signal k 1 Y (s) = t s +1 s c. we use partial fractions expansion and then inverse Laplace transform to obtain the system response in time domain A B Y (s) = + t s +1 s k k A = Y ( s )(t s + 1) s =- 1 = = -t k and B = Y ( s ) s s =0 = =k s s =- 1 t s + 1 s =0 t

993 thus at t = 5t the system step response reaches more than 99% of the steady state value. one can determine the time constant of the system using
(t2 -t1 ) y (t1 ) t . i. the DC gain. a b. t2 . in order to determine the value of the
t ®¥
system’s DC gain one needs to divide the measured steady state value by the amplitude of the step input. c. is k -t y (t ) = e t u-1 (t ) t Measuring the value of the impulse response at positive time moments. In order to determine the value of k one notes that the steady state value of the step response is lim y (t ) = k which can be easily measured from the step response graph. Then the steady state value of the system’s step
response is lim y (t ) = ak .p is the pole t s +1 s + 1/ t s + p of the system.This equation gives the value of the time constant of the system if y (t2 ). k k /t q e.e-1 ) = k * 0. considering initial conditions equal with 0. d. which means that the
t ®¥ s ®0
gain of the system at zero frequency. say U ( s ) = where a denotes the s amplitude of the step input. The impulse response of this system. k are known. If the step input is not of amplitude 1. then the step response of the system is y (t ) = a (. y (t1 ). Let t = 5t then y (t ) = k (1 . 1 and the relations given above to verify that the step response 2 presented in the figure is indeed the step response of a system described by G ( s ) = .e-5 ) = k * 0. In this case. is k . =e y (t2 ) 1 1
3
. G ( s ) = can also be written as G ( s ) = = where . t1. Let t = t then y (t ) = k (1 . 3s + 1 Notes: a.
t ®¥
Question 1: Use the graph in Fig.6321 thus the system’s time constant is the time moment when the system step response reaches approximately 63% of the steady state value.e. Remember also the final value theorem lim y (t ) = lim Y ( s ) . say t1 and t2 .ke
-t 1 t
+ ku-1 (t )) .

a . wn which defines it as
5
.II.
s ®0
The denominator is the Characteristic polynomial which can be written in several canonical forms. The norm of the vector from the origin to the pole is wn. Figure 3 shows the location of the two poles in the s-plane. The real part of the poles is -a and the imaginary part is jb. which is known as the natural frequency. One may write 2 D( s ) = s 2 + 2as + w n = ( s + a ) 2 + b 2 where 2 b 2 +a 2 = wn . wn ) . Second order system (complex pole pair) A second order system with no finite zeros is described in frequency domain by the transfer function kw 2 k0wn 2 k0wn 2 H ( s) = 2 0 n = º D( s) s + 2a s + wn 2 s 2 + 2zwn s + wn 2 The numerator is chosen to scale the transfer function so that the DC gain (which can be calculated by lim H ( s ) ) is equal to k0 . Location of the complex poles of a second order system in the s-plane Notice that the system dynamics is completely described by the triplets (k0 .
2
Figure 3. Thus if a 2 < w n . the polynomial D( s ) describes a complex pair of poles at s = -a ± jb . z . To link the two forms of the characteristic polynomial we define the damping ratio as a z = . b ) or ( k 0 . including 2 2 D( s ) = s 2 + 2as + w n = s 2 + 2zw n s + w n .

If z = 1 then b = 0.
The impulse response of the system with transfer function kw 2 H ( s) = 2 0 n (s + a )2 + b 2 is given by wn 2 -a t y (t ) = k0 e sin b t u-1 (t ) .
2
6
. If 0 > z > -1 then one has a complex pair in the right-half plane (e. b = w n and the poles are at s = ± jb on the imaginary axis. Note that one may write z = cos q = . both at s = -a . This is known as the underdamped case. the period of the oscillation is given by T 2p T= .cos j = b = wn -a 2 = wn 1-z 2 . b The variable b is known as the oscillation frequency. which provides the exponential decay term.
2
Figure 4. This is known as the undamped case. In this overdamped case.a = w n and the poles are on the real axis. Having in mind the standard form for sinusoids sin 2p t . The figure clearly shows the meaning in the time domain of the real part -a of the poles. wn For complex poles in the left-half plane one has 0 < z < 1 .g. If z > 1 then there are two real poles and we can split the quadratic factor D( s ) = s 2 + 2as + w n into two real linear factors. Sketch of the impulse response of a second order system with complex poles If z = 0 then a = 0. unstable complex pair).a . b which is plotted for 0 < z < 1 in the figure. the impulse response has the form te -at .

4 s + 1
which have the same time constant t = 0. then s y yss = ak0 and thus k0 = ss .4 . a Note that in the case in which the system input is not a unit step. a In Figure 6 are compared the step responses of the two systems H1 ( s ) = H 2 (s) = 6 s + 5s + 6
2
1 and 0. from the system’s step response. one can simply calculate the POV and then determine the damping factor z . In this case however the signal rises faster and one may 2. Measuring the steady state value of the step response one gets the system DC gain k0 = yss . U ( s ) = .8 . tr. using t r = wn The percent overshoot is a function of damping ratio POV = 100e -p z / and conversely
1-z 2
ù é ln 2 POV 100 ú z =ê 2 ê ln POV +p 2 ú 100 û ë
(
(
)
)
1/ 2
.
To determine the parameters of the system. Then by measuring the settling time one can use t s = 5t to determine the time constant of the system and then a . for 0. is the time required for the step response to rise from 0. (Notice the difference.16z + 0. Note that for a first order system one has t r = 2.An important quantity for characterizing the performance of systems is the percent overshoot (POV) in the step response. This is defined as y .)
8
. The settling time ts is the time required for the signal to effectively reach its steady-state value.2t and t s = 5t (some take t s = 4t ). the time constant is t = 1 and one may use t s = 5t to a calculate the settling time.g. The rise time. e.9 of its steady-state value. For the underdamped pole pair. approximate.6 .1 to 0.3 £ z £ 0.yss POV = max ´ 100% yss where ymax is the maximum value of the step response and yss is its steady-state value.

5
Figure 6.5 0.
9
.5
1
1.1 0 second order system first order system
0
0.4 0.7 0.8 0.Step Response 1 0.6 Amplitude 0. Step response comparison of a first order system and a second order. overdamped.5
3
3.9 0.5 Time (sec)
2
2. system. Both systems have the same time constant thus the same settling time.3 0.2 0.

717 System: x Time (sec): 2. Determine the model transfer function of the system.994
0.2.32 Amplitude: 1.5 Time (sec)
2
2.624 Amplitude: 0.53
1. To help with the calculations.5
1
1.2
System: x Time (sec): 1.2
0
0
0. some points were given explicitly on the graph.95 System: x Time (sec): 0.6
0.59 Amplitude: 1.4
1. Step response of a system
Consider that the steady state is obtained when the signal enters in the ±1% of the steady state value (this is related with the value of the settling time). The following figure presents the step responses (to a unit step) of a system.6 System: x Time (sec): 0.35 Amplitude: 0.
Step Response 1.5 Amplitude: 0.04
1 Amplitude System: x Time (sec): 1.5
3
Figure 3.
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.4
0.8
0.